If I have a 9Vp-p with 12khz going through a 4.7k resistor how do i figure what the Vp-p value that the resistor drops
The question you have asked cannot be answered unless you describe the circuit more completely. The best way would be to show a diagram. If the resistor is connected straight across the 9Vp-p supply and common with no other parts intervening then the answer would be 9Vp-p. If there is more to the circuit, the answer may be different. It is not possible to say how much voltage a resistor will drop when it is connected to a supply, without also knowing what load it is connected to.
9Vp-p -------4.7kresistor--------1μF capacitor-----1μF cap.------10kΩ 12khz Then you have a 15Vdc in series with a 3.3kΩ and a 5.6kΩ and that is running between the capacitors. would like to find out the Vp-p right after the first resistor.
The circuit structure is still not clear: really it would be much better if you could post a diagram. (What does "running between the capacitors" mean - in series with them? in parallel with them?) Is the circuit like the one in this thread? http://forum.allaboutcircuits.com/showthread.php?t=64346 Note that in any case the reactance of 1μF capacitors at 12kHz is only about 13Ω, which is relatively small compared to the resistor values quoted, so as far as AC is concerned you could probably consider the capacitors as short-circuits, unless you need to calculate very exact amplitude levels, or a value for the phase shift.